Question. Proof of divides - 2 approaches












2












$begingroup$


Suppose $n$ is an integer. If 3|n then 3|$n^2$. Prove.



So I'm wondering if both approaches here are ok.



$1^{st}$:



3|n so $n=3a$ a in integers



$n^2=9a^2$



$n^2=3(3a^2)$



$2^{nd}$:



$n=3a$



(both cases we have 3 times an integer) Thanks



$n^2=3an$






proof-writing







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Both approaches are valid, yeah.
    $endgroup$
    – Eevee Trainer
    Dec 4 '18 at 22:10










  • $begingroup$
    I wouldn't say there is really any difference between these two approaches.
    $endgroup$
    – Morgan Rodgers
    Dec 4 '18 at 23:24
















2












$begingroup$


Suppose $n$ is an integer. If 3|n then 3|$n^2$. Prove.



So I'm wondering if both approaches here are ok.



$1^{st}$:



3|n so $n=3a$ a in integers



$n^2=9a^2$



$n^2=3(3a^2)$



$2^{nd}$:



$n=3a$



(both cases we have 3 times an integer) Thanks



$n^2=3an$






proof-writing







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Both approaches are valid, yeah.
    $endgroup$
    – Eevee Trainer
    Dec 4 '18 at 22:10










  • $begingroup$
    I wouldn't say there is really any difference between these two approaches.
    $endgroup$
    – Morgan Rodgers
    Dec 4 '18 at 23:24














2












2








2





$begingroup$


Suppose $n$ is an integer. If 3|n then 3|$n^2$. Prove.



So I'm wondering if both approaches here are ok.



$1^{st}$:



3|n so $n=3a$ a in integers



$n^2=9a^2$



$n^2=3(3a^2)$



$2^{nd}$:



$n=3a$



(both cases we have 3 times an integer) Thanks



$n^2=3an$






proof-writing







share|cite|improve this question











$endgroup$




Suppose $n$ is an integer. If 3|n then 3|$n^2$. Prove.



So I'm wondering if both approaches here are ok.



$1^{st}$:



3|n so $n=3a$ a in integers



$n^2=9a^2$



$n^2=3(3a^2)$



$2^{nd}$:



$n=3a$



(both cases we have 3 times an integer) Thanks



$n^2=3an$






proof-writing




proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 23:10









Crabzmatic

302214




302214










asked Dec 4 '18 at 22:09









HarryHarry

253




253








  • 5




    $begingroup$
    Both approaches are valid, yeah.
    $endgroup$
    – Eevee Trainer
    Dec 4 '18 at 22:10










  • $begingroup$
    I wouldn't say there is really any difference between these two approaches.
    $endgroup$
    – Morgan Rodgers
    Dec 4 '18 at 23:24














  • 5




    $begingroup$
    Both approaches are valid, yeah.
    $endgroup$
    – Eevee Trainer
    Dec 4 '18 at 22:10










  • $begingroup$
    I wouldn't say there is really any difference between these two approaches.
    $endgroup$
    – Morgan Rodgers
    Dec 4 '18 at 23:24








5




5




$begingroup$
Both approaches are valid, yeah.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 22:10




$begingroup$
Both approaches are valid, yeah.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 22:10












$begingroup$
I wouldn't say there is really any difference between these two approaches.
$endgroup$
– Morgan Rodgers
Dec 4 '18 at 23:24




$begingroup$
I wouldn't say there is really any difference between these two approaches.
$endgroup$
– Morgan Rodgers
Dec 4 '18 at 23:24










1 Answer
1






active

oldest

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0












$begingroup$

Both are correct. The 2nd is essentially $ 3mid nmid nm,Rightarrow,3mid nm $ by transitivity of "divides" (OP is case $,m = n)$. Thus the set $,3Bbb Z,$ of multiples of $3$ is closed under multiplication by any integer $m$. Similarly you can show it is closed under addition (so also subtraction by $,a-b = a+(-1)b,]$. Therefore $, 3mid m,n,Rightarrow, 3mid j m + k n,$ for all integers $,j,k. $ Ditto if we replace $3$ by any integer.






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    $begingroup$

    Both are correct. The 2nd is essentially $ 3mid nmid nm,Rightarrow,3mid nm $ by transitivity of "divides" (OP is case $,m = n)$. Thus the set $,3Bbb Z,$ of multiples of $3$ is closed under multiplication by any integer $m$. Similarly you can show it is closed under addition (so also subtraction by $,a-b = a+(-1)b,]$. Therefore $, 3mid m,n,Rightarrow, 3mid j m + k n,$ for all integers $,j,k. $ Ditto if we replace $3$ by any integer.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Both are correct. The 2nd is essentially $ 3mid nmid nm,Rightarrow,3mid nm $ by transitivity of "divides" (OP is case $,m = n)$. Thus the set $,3Bbb Z,$ of multiples of $3$ is closed under multiplication by any integer $m$. Similarly you can show it is closed under addition (so also subtraction by $,a-b = a+(-1)b,]$. Therefore $, 3mid m,n,Rightarrow, 3mid j m + k n,$ for all integers $,j,k. $ Ditto if we replace $3$ by any integer.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Both are correct. The 2nd is essentially $ 3mid nmid nm,Rightarrow,3mid nm $ by transitivity of "divides" (OP is case $,m = n)$. Thus the set $,3Bbb Z,$ of multiples of $3$ is closed under multiplication by any integer $m$. Similarly you can show it is closed under addition (so also subtraction by $,a-b = a+(-1)b,]$. Therefore $, 3mid m,n,Rightarrow, 3mid j m + k n,$ for all integers $,j,k. $ Ditto if we replace $3$ by any integer.






        share|cite|improve this answer











        $endgroup$



        Both are correct. The 2nd is essentially $ 3mid nmid nm,Rightarrow,3mid nm $ by transitivity of "divides" (OP is case $,m = n)$. Thus the set $,3Bbb Z,$ of multiples of $3$ is closed under multiplication by any integer $m$. Similarly you can show it is closed under addition (so also subtraction by $,a-b = a+(-1)b,]$. Therefore $, 3mid m,n,Rightarrow, 3mid j m + k n,$ for all integers $,j,k. $ Ditto if we replace $3$ by any integer.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 4 '18 at 22:38

























        answered Dec 4 '18 at 22:33









        Bill DubuqueBill Dubuque

        209k29190632




        209k29190632






























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